What is the following limit?

lim as n goes to infinity of (pi/n) (sin(pi/n) + sin(2pi/n) + sin(3pi/n) +...+ sin(npi/n)) =
I.) lim as n goes to infinity sigma (n and k=1) of pi/n sin(kpi/n)
II.) Definite integral from 0 to pi of sin(x)dx
III.) 2

A.) I only
B.) II only
C.) III only
D.) II and III only
E.) I, II, and III

Wow, I am really lost on this one, please help!

To find the limit of the given expression, let's break it down step by step:

I.) First, let's rewrite the given expression as a sum using the sigma notation:
lim as n goes to infinity of (pi/n) (sin(pi/n) + sin(2pi/n) + sin(3pi/n) +...+ sin(npi/n)) = lim as n goes to infinity of (pi/n) * Σ (k from 1 to n) sin(kpi/n).

II.) Next, notice that the term inside the limit resembles the Riemann sum for the definite integral ∫[a, b] f(x)dx, where f(x) = sin(x) and [a, b] = [0, pi]. The Riemann sum is given by Σ (k from 1 to n) f(x) * Δx, where Δx = (b-a)/n.

In this case, we can let a = 0 and b = pi, so Δx = (pi-0)/n = pi/n. This means that our sum simplifies to Σ (k from 1 to n) sin(kpi/n) * (pi/n).

III.) Finally, let's take a look at the definite integral of sin(x) from 0 to pi. Using the fundamental theorem of calculus, we know that ∫[a, b] f(x)dx = F(b) - F(a), where F(x) is the antiderivative of f(x).

In this case, the antiderivative of sin(x) is -cos(x), so the definite integral from 0 to pi of sin(x)dx evaluates to -cos(pi) - (-cos(0)) = -(-1) - (-1) = 2.

Putting it all together, we can conclude that the limit of the given expression is equivalent to the definite integral of sin(x) from 0 to pi, which is equal to 2.

Therefore, the correct answer is:
C.) III only